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Sine equations

The equation sin⁡x=a\sin x = a has two solutions during one revolution, as long as aa lies between −1-1 and 11. One is the inverse sine of aa, and the other is π\pi minus the first. Then you can add whole revolutions, n⋅2πn\cdot 2\pi. For a=1/2a = 1/2 the first solution is π/6\pi/6. You should check both by inserting them into sine.

x=sin⁡−1a+n⋅2πx = \sin^{-1} a + n\cdot 2\pifirst family
x=π−sin⁡−1a+n⋅2πx = \pi - \sin^{-1} a + n\cdot 2\pisecond family

Symbols

aathe right-hand side, between -1 and 1
nnan integer

Example

sin⁡x=12\sin x = \dfrac{1}{2} gives x=π6x = \dfrac{\pi}{6}, where sine is 0.50.5, and x=π−π6≈2.618x = \pi - \dfrac{\pi}{6} \approx 2.618.

Both solutions in the interval [0,2π)[0, 2\pi) should be included, not only the one from the calculator.
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Part of Mathematics R2: Trigonometric functions.